How do you find the vertex, focus and directrix of #x+y^2=0#?

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Answer 1 · 2018-08-03T12:42:16

Vertex is at #(0,0)#, focus is at #(-0.25,0)# and
directrix is #x= 0.25#

# x+y^2=0 or y^2 = -x or (y-0)^2 = -4 *1/4 (x-0)#

The equation of horizontal parabola opening left is

#(y-k)^2 = -4 p(x-h) ; h=0 ,k=0 ; p= 1/4=0.25#

Therefore, vertex is at #(0,0)# The distance between focus and

vertex is #p=0.25 :.# Focus is at #(-0.25,0)# . Vertex is at midway

between focus and directrix. Therefore , directrix is #x= 0.25#

graph{x+y^2=0 [-12.66, 12.65, -6.33, 6.33]} [Ans]